Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

4. Then, if only one difference stands against any rate, it will be the quantity belonging to that rate, but if there be ac veral, their sum will be the quantity.

[ocr errors]

d.

12

Ansiccr.

EXAMPLES. 1. A merchant has spices, some at 90. per lb. some at Is. some at 2s. and some at 2s. 6d. per Ib. how much of cach sort must he mix, that he may sell the inixture at Is. 8d. per pound ? d. d.

lb. 9

10 at 97 d.

4 12 [ Gives the d. 12 10 2024 8 24 Answer ; or

2024) 11 30 11 30

30

8 2. A grocer would mix the following qualities of sugar; viz. at 10 cents, 13 cents, and 16 cents per Ib. ; what

lantity of each sort must be taken to make a mixture worth 12 cents per pound? Ans. 5 lb. at 10 cts. 2 lb. at 13 cts. and 2 lb. at 16 cts. per lb.

3. A grocer has two sorts of tea, viz. at 9s. and at 159. per Ib. how must he inix them so as to afford the composition for 12s.

per

Ib. ? Ans. He must mir an equal quantity of each sort. 4. A goldsmith would mix gold of 17 carats finc, with some of 19, 21, and 24 carats fine, so that the compound inay be 22 carats fine; what quantity of each must he take?

Ans. 2 of each of the first thrce sorts, and 9 of the last. 5. It is required to mix several sorts of rum, viz. at 5s. 7s. and Is. per gallon, with water at O per gallon, together, so that the mixture inay be worth 6s. per gallon; how much of each sort must the mixture consist of? Ans. I gal. of rum at 58., 1 do. at 7s., 6 do. at 9s. and 3 gals. water. Or, 3 gals. rum at 58., 6 do. at 7s., I do. at 9s. and

1 gal. water. 6. A grocer hath several sorts of sugar, viz. one sort at 12 cts. per lb. another at 11 cts. a third at 9 cts. and a fourth at 8 cts. per Ib. ; I demand how much of each sort he must mix together, that the whole quantity may be afforded at 10 cents per pound?

[ocr errors]

1b. cts.
16. cts,

lb. cts. 52 at 12

1 at 12

53 at 12 1 at 11

2 at 11 1st Ans.

2d Ans.

3d Ans.

| 2 at 11 I at 9

2 at
9

2 at 9 2 at 8

1 at
8

3 at 8 4th Ans. 3 lb. of each sort.*

CASE II.

ALTERNATION PARTIAL, Or, when one of the ingredients is limited to a certain quantity, thence to find the several quantities of the rest, in proportion to the quantity given.

RULE. Take the differences between each price, and the mean rate, and place them alternately as in CASE I. Then, as the difference standing against that simple whose quantity is given, is to that quantity : so is each of the other differ ences, severally, to the several quantities required.

EXAMPLES,

1. A farmer would mix 10 bushels of wheat, at 70 cents per busbel, with rye at 48 cts. corn at 36 cts. and barley at 30 cts. per bushel, so that a bushel of the composition may be sold for 38 cts.; what quantity of each must he iaken?

70 8 stands against the given quan48 2

(tity. Mean rate, 38

36 10
30

32

2 : 2 bushels of rye. As 8 : 10 :: 10 : 12; bushels of corn.

32 : 40 bushels of barley.

[ocr errors]

* These four answers arise from as many various ways of linking the rates of the ingredients together.

Questions in this rule admit of an infinite variety of answers: for after the quantities are found from different methods of linking; any other numbers in the same proportion between themselves, as the numbers which compose the answer, will likewise satisfy the conditions of the question.

2. How much water must be mixed with 100 gallons of "rum, worth 7s. 6d. per gallon, to reduce it to 6s. 3d. per gallon?

Ans. 20 gallons. 3. A farmer would mix 20 bushels of rye, at 65 cents per bushel, with barley at 51 cts. and oats at 30 cents per bushel; how much barley and oats must be mixed with the 20 bushels of rye, that the provender may be worth 41 cts. per bushel ?

Ans. 20 bushels of barley, and 61 | bushels of oats. 4. With 95 gallons of rum at 8s. per gallon, I mixed other rum at 6s. 8d. per gallon, and some water; then I found it stood me in Os. 4d. per gallon; I demand how much rum and how much water I took ?

Ans. 95 gals. rum at Cs. 8d. and 30 gals. water.

CASE III.

When the whole composition is limited to a given quantity.

RULE.

Place the difference between the mean rate, and the several prices alternately, as in Case I. ; then, As the sum of the quantities, or difference thus determined, is to the given quantity, or whole composition: so is the difference of each rate, to the required quantity of each rate.

EXAMPLES.

1. A grocer had four sorts of tea, at Is. 3s. 6s. and 10s. per lb. the worst would not sell, and the best were too dear; he therefore mixed 120 lb. and so much of each sort, as to sell it at 4s. per lb.; how much of each sort did he take? 1 6

76:60 at 1 3 2 Ib. lh.

; 2: 20 3
6

1 As 12 : 120 : : 1:10 6
3

3:30 - 10

per lb.

[blocks in formation]

2. How much water at 0 per gallon, must be mixed witlo wine at 90 cents per gallon, so as to fill a vessel of 100 gallons, which may be afforded at 60 cents per gallon ?

Ans. 33} gals. water, and 66; gals. wine. 3. A grocer having sugars at 8 cts. 16 cts. and 24 cte. per pound, would make a composition of 240 lb. worth 20 cts. per lb. without gain or loss; what quantity of each must be taken ? Ans. 40 lb. at 8 cts. 40 lb. at 16 cts. and 160 lb, at 24 cts.

4. A goldsmith had two sorts of silver bullion one of 10 oz. and the other of 5 oz. fiue, and has a mind to mix a pound of it so that it shall be 8 oz. fine; how much of each sort must he take?

Ans. 1 of 5 oz. fine, and 7} of 10 oz. fine. 5. Brandy at 3s. 6d. and 5s. Id. per gallon, is to be mixed, so that a hhd. of 63 gallons may be sold for 121. 12s.; how many gallons must be taken of each?

Ans. 14 gals. at 5s. 9d. and 49 gals. at 3s. 6d.

ARITHMETICAL PROGRESSION. ANY rank of numbers more than two, increasing by common excess, or decreasing by common difference, is said to be in Arithmetical Progression.

2,4,6,8, &c. is an ascending:?ritlimetical series : So

8,6,4,2. &c. is a descending arithmetical series : The numbers which form the series, are caller the terra of the progression ; the first and last terms of which ara called the extremes.*

PROBLEM I.

The first term, the last term, and the number of terms being given, to find the sum of all the terms.

* A series in progression includes five parts, viz. the first term, last term, number of terms, common differ pce, and sun of the series.

By having any three of the parts given, the other two may be found which admits of a variety of Problems; but most of them are best under stood by an algebraic process, and are here omitted.

ROLE. --Multiply the sum of the extremes by the nui. terms, and half the product will be the answer.

EXAMPLES.

1. The first term of an arithmetical series is 3, the last term 23, and the number of terms 11; required the sum of the series.

23-4-3=26 sum of the extremnes.

Then 2611=2=143 the Answer. 2. How many strokes does the hammer of a clock strike in 12 hours.

Ans. 78. 3. A merchant sold 100 yards of cloth, viz. the first yard for 1 ct. the second for 2 cts. the third for 3 cts. &c. I demand what the clotlı came to at that rate ?

Ans. $501 4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave Is. and for the last yd. 11. 178. what did the whole come to?

Ans. £18 ls. 5. A draper sold 100 yards of broadcloth, at 5 cts. for the first yard, 10 cts. for the second, 15 for the third, &c. increasing 5 cents for every yard; what did the whole amount to, and what did it average per yard ?

Ans. Amount $252, and the average price is $2, 52 cts. 5 mills per yard.

6. Suppose 144 oranges were laid 2 yards distant from each other, in a right line, and a basket placed two yards from the first orange, what length of ground will that boy travel over, who gathers them up singly, returning with them one by one to the basket ?

Ans. 23 miles, 5 furlɔngs, 180 yds.

PROBLEM II.

The first term, the last term, and the number of terms given,

to find the common difference.

RULE.--Divide the difference of the extremes by the number of terms less ), and the quotient will be the common difference.

« ΠροηγούμενηΣυνέχεια »